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aceaceBest ResponseYou've already chosen the best response.0
Find a relation between x and y that does NOT involve logarithms: log x + log y = log (x+ y)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Can we assume it's a log of base 10?
 one year ago

AERONIKBest ResponseYou've already chosen the best response.0
logx + log y is never log(x+y) it seems... it is log(xy)
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
do you want the answer?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
I suppose we could exponentiate both sides.. \[\huge 10^{(\log x + \log y)}=10^{\log(x+y)}\] Using rules of exponents we can split up the left side like this: \[\huge 10^{\log x}*10^{\log y}=10^{\log(x+y)}\] Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". Giving us: \[\huge xy=(x+y)\] \[\huge y=\frac{x}{x1}\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Something like that maybe? :o
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
@zeptr yes thats the answer !!!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Were you just testing us or something? :D lol Knew the answer the whole time huh? XD
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
@AERONIK , it seems that logx+logy=log(x+y) sometimes. I wouldn't say NEVER. Obviously it doesn't match the rule for logs that we had in mind. But if you plug in y=2, x=2 into the original equation, that should work out ok as one solution :D Since 2+2 is the same as 2*2
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
no i had the answers at the back of book but i have no idea how to do it... i still dont get the logic though...
 one year ago

AERONIKBest ResponseYou've already chosen the best response.0
so it is not an identity and just a function you say ?
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
yeh just a random equation
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Yah it appears it wasn't testing you on whether or not you knew your log identities, they just wanted to see a relationship between x and y. Interesting problem :O
 one year ago

AERONIKBest ResponseYou've already chosen the best response.0
and one more thing in your proof you have assumed the given statement to be correct, can you please give a formal proof for te same?
 one year ago

AERONIKBest ResponseYou've already chosen the best response.0
and is it log to the base 10 or log to the base e ?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
For this problem, we have to assume that they are ALL the same base. Otherwise it won't work out the same. If they're a different base, let's say A, then we exponentiate, writing each side with a base A (regardless of what A might be) and it will work out the same! :D
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
Different than 10 i mean*
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
10 logx ∗10 logy =10 log(x+y) Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". HOW DID YOU DO THIS STEP... CAN YOU SHOW ME? PLEASE
 one year ago

AERONIKBest ResponseYou've already chosen the best response.0
by defination if y=log x to the base a, then x is nothing but a^y, just use this and you can prove the identity.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
\[\large \arcsin( \sin x)=x\]\[\huge e^{\ln x}=x\] Yah this is true of any function and it's inverse :D Remember it in trig with the inverse functions?
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
no ...can you do a proof for it please?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.3
dw:1353225574033:dw Hmmm this might be a little confusing.. I drew a lot of arrows : lemme know if it makes some sense.
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
i get it ...thanks so much
 one year ago
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